1. Field of the Invention
The present invention relates to a reference signal generator circuit for use in, e.g., a wireless communication system, and in particular, a reference signal generator provided with two 90-degree phase shifters and two mixer circuits.
2. Description of the Related Art
In MB-OFDM (Multi-Band Orthogonal Frequency Division Multiplexing) which is a scheme for a UWB (Ultra Wide Band) system, as shown in FIG. 8, a frequency band (a band) in a range from 3.1 GHz to 10.6 GHz is divided into five groups G1 to G5. Each of the groups G1 to G5 has three sub-frequency bands (subbands) and requires high-rate frequency switching of 9.2 nanoseconds. Conventionally, oscillators for handling the seven subbands (B1 to B7 and B8 to B14) have been proposed in such a manner that wide-band PLL circuits are switched or a plurality of SSB mixers are used.
FIG. 9 is a circuit diagram showing a configuration of a conventional reference signal generator circuit for use in a wireless communication system. FIG. 10 is a spectrum diagram showing a reference signal generated by the reference signal generator circuit shown in FIG. 9. Referring to FIG. 9, the reference signal generator circuit is configured to include a center frequency oscillator 90 for generating a reference signal having a center frequency fc, mixers 91 and 92, a shift frequency oscillator 93 for generating a local oscillation signal having a shift frequency Δf, a 90-degree phase shifter 94, low pass filters (LPFs) 95 and 97, high pass filters (HPFs) 96 and 98, and multiplexers 99 and 100 for selectively switching between two input signals in accordance with a selection signal Ss to output reference signals S1 and S2.
In the reference signal generator circuit configured as described above, the mixers 91 and 92 output mixed signals M1 and M2, respectively, and the mixed signals M1 and M2 are expressed by the following equations:
                                                                        M                ⁢                                                                  ⁢                1                            =                                                sin                  ⁡                                      (                                          2                      ⁢                                              πΔ                        ⁢                        ft                                                              )                                                  ×                                  cos                  ⁡                                      (                                          2                      ⁢                      π                      ⁢                                                                                          ⁢                      fct                                        )                                                                                                                          =                                                (                                      1                    /                    2                                    )                                ⁢                                  {                                                            sin                      ⁡                                              (                                                  2                          ⁢                                                      π                            ⁡                                                          (                                                              fc                                +                                                                  Δ                                  ⁢                                                                                                                                          ⁢                                  f                                                                                            )                                                                                ⁢                          t                                                )                                                              -                                          sin                      ⁢                                              (                                                  2                          ⁢                                                      π                            ⁡                                                          (                                                              fc                                -                                                                  Δ                                  ⁢                                                                                                                                          ⁢                                  f                                                                                            )                                                                                ⁢                          t                                                )                                                                              }                                                                                                                                          =                                                            (                                              1                        /                        2                                            )                                        ⁢                                          (                                                                        H                          ⁢                                                                                                          ⁢                          1                                                -                                                  L                          ⁢                                                                                                          ⁢                          1                                                                    )                                                                      ,                            ⁢                                                                                                      (        1        )                        and                                                                                                          M                ⁢                                                                  ⁢                2                            =                                                cos                  ⁡                                      (                                          2                      ⁢                                              πΔ                        ⁢                        ft                                                              )                                                  ×                                  cos                  ⁡                                      (                                          2                      ⁢                      π                      ⁢                                                                                          ⁢                      fct                                        )                                                                                                                          =                                                (                                      1                    /                    2                                    )                                ⁢                                  {                                                            cos                      ⁡                                              (                                                  2                          ⁢                                                      π                            ⁡                                                          (                                                              fc                                +                                                                  Δ                                  ⁢                                                                                                                                          ⁢                                  f                                                                                            )                                                                                ⁢                          t                                                )                                                              +                                          cos                      ⁡                                              (                                                  2                          ⁢                                                      π                            ⁡                                                          (                                                              fc                                -                                                                  Δ                                  ⁢                                                                                                                                          ⁢                                  f                                                                                            )                                                                                ⁢                          t                                                )                                                                              }                                                                                                                        =                                ⁢                                                      (                                          1                      /                      2                                        )                                    ⁢                                      (                                                                  H                        ⁢                                                                                                  ⁢                        2                                            +                                              L                        ⁢                                                                                                  ⁢                        2                                                              )                                                              ,                                                          (        2        )            
where, H1 and H2 denote upper side band signals outputted from the HPFs 96 and 98, respectively, and L1 and L2 denote lower side band signals outputted from the LPFs 95 and 97, respectively. The multiplexers 99 and 100 selectively switch between the two input signals to output reference signals S1 and S2 called an I signal and a Q signal which are different in phase from each other by 90 degrees.
Documents related to the present invention are as follows:    (a) Patent Document 1: Japanese Patent Laid-open Publication No. 2004-523142;    (b) Patent Document 2: Japanese Patent Laid-open Publication No. 2007-295068;    (c) Patent Document 3: U.S. Pat. No. 7,321,268; and    (d) Non-patent Document 1: Geum-Young Tak et al., “A 6.3-9-GHz CMOS Fast Setting PLL for MB-OFDM UWB Applications”, IEEE Journal of Solid-state Circuits, Vol. 40, No. 8, August 2005.
However, the use of the wide-band PLL circuit requires control of wide-band frequencies for the plurality of groups and control of fine frequencies corresponding to the subbands. Consequently, this case makes it considerably difficult to cover all the groups in the MB-OFDM (e.g., See Non-patent Document 1). In addition, when a normal mixer is used, a filter is required (e.g., See FIG. 9). Alternatively, this case requires four or more SSB mixers (e.g., See Patent Document 2). Hence, there is a problem of increase in circuit scale.
In addition, the case where the filters 95 to 98 are used in accordance with the conventional example shown in FIG. 9 causes the following problem. For example, if the shift frequency Δf is relatively small, a signal from one filter can not be completely separated from a signal from another filter with ease, as shown in FIG. 10.